leetcode1004

 class Solution:
     def getMax(self,B:'List[int]'):
         n = len(B)
         maxlen = 0
         curlen = 0
         for i in range (n):
             if B[i] == 1:
                 curlen += 1
             else:
                 maxlen = max(maxlen,curlen)
                 curlen = 0
         return max(maxlen,curlen)
     def longestOnes(self, A: 'List[int]', K: int) -> int:
         if K == 0:
             return self.getMax(A)
         n = len(A)
         zlist = list()
         for i in range(n):
             if A[i] == 0:
                 zlist.append(i)
         if len(zlist)<=K:
             return n
         maxlen = 0
         for zi in range(len(zlist)-K+1):
             ti = zi+K
 
             left = 0
             right = len(A)-1
             if zi==0:
                 left = zlist[zi]
             else:
                 left = zlist[zi-1]+1
             if zi == len(zlist)-K:
                 right = len(A) - 1
             else:
                 right = zlist[ti] - 1
 
             maxlen = max(maxlen, right - left +1)
 
         return maxlen

经过了几次尝试，终于作出来了。主要的思路是滑动窗口：

先记录所有的0的索引，然后选择等K宽的窗口，计算窗口“所连接”的连续1的起止坐标。然后滑动窗口，进行比较，保留最大值。